Name the source of lactose and name the type of enzyme that is required to digest it

Chemistry

1 answer:

skelet666 [1.2K]3 years ago

3 0

Answer:

Lactose is found in animal milk is lactase is the enzyme that is required to digest it.

Explanation:

Lactose is a disaccharide made of glucose and galactose units. This sugar is naturally present in animal milk and it is digested (broken in its units) by the lactase enzyme.

Some people are lactose intolerant because their bodies is not able to produce the enzyme. If they ingest dairy products it may cause health issues. Nowadays it is possible to purchase the lactase enzyme and ingest it with a dairy food to avoid any health effects.

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An acidic fog in pasadena was found to have a ph of 2.50. which expression represents this ph measurement?

joja [24]

Supposing a temperature of 25 degrees and supposing that all activity coefficients are 1

pH = -log[H+]
pOH = -log[OH-]
pH + pOH = 14

Thus a pH of 2.50 would mean that the [H+], the concentration of the hydrogen ion, would be 10^(-2.50)

pH + pOH = 14
pOH = 14 - pH = 14 - 2.5 = 11.5
MOH- levels would be coordinated with pOH
pOH = -log[OH-] ==> [OH-] = [MOH-] = 10^-pOH = 10^-11.5 = 3.2 x 10^-12

Therefore, MOH¯ = 3.2 × 10¯12 M

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3 years ago

Calculate the OH− concentration after 53 mL of the 0.100 M KOH has been added to 25.0 mL of 0.200 M HBr. Assume additive vol- um

Andre45 [30]

Answer:

$\large \boxed{\text{0.0038 mol/L}}$

Explanation:

1. Calculate the initial moles of acid and base

$\text{moles of acid} = \text{0.0250 L} \times \dfrac{\text{0.200 mol}}{\text{1 L}} = \text{0.005 00 mol}\\\\\text{moles of base} = \text{0.053 L} \times \dfrac{\text{0.100 mol}}{\text{1 L}} = \text{0.0053 mol}$

2. Calculate the moles remaining after the reaction

OH⁻ + H₃O⁺ ⟶ 2H₂O

I/mol: 0.0053 0.005 00

C/mol: -0.00500 -0.005 00

E/mol: 0.0003 0

We have an excess of 0.0003 mol of base.

3. Calculate the concentration of OH⁻

Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L

$\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}$